On the Path Sequence of a Graph
S lawomir Bakalarski, Jakub Zygad lo Institute of Computer Science and Computational Mathematics Faculty of Mathematics and Computer Science, Jagiellonian University
ul. Lojasiewicza 6, 30-348 Krak´ow, Poland
e-mail: Slawomir.Bakalarski@uj.edu.pl, Jakub.Zygadlo@ii.uj.edu.pl
Abstract. A subset S of vertices of a graph G = (V, E) is called a k-path vertex cover if every path on k vertices in G contains at least one vertex from S. Denote by ψk(G) the minimum cardinality of a k-path vertex cover in G and form a sequence ψ(G) = (ψ1(G), ψ2(G), . . . , ψ|V |(G)), called the path sequence of G.
In this paper we prove necessary and sufficient conditions for two integers to appear on fixed positions in ψ(G). A complete list of all possible path sequences (with multiplicities) for small connected graphs is also given.
Keywords: k-path vertex cover, path sequence, list for small graphs.
1. Introduction
Let G be a graph and let k be a positive integer. Following [3], define a k-path vertex cover (k-PVC for short) of G as a subset S of vertices of G such that every path on k vertices in G has at least one vertex in common with S. A k-PVC is called minimum if it has minimum cardinality among all k-path vertex covers of G. This minimum cardinality is denoted by ψk(G) and called a k-path number of G. Our main subject is to investigate the properties of path sequences, defined as follows:
Definition 1 Let G = (V, E) be a graph on n vertices. The sequence of all path numbers, namely
ψ(G) = (ψ1(G), ψ2(G), . . . , ψn(G)) will be called a path sequence of G.
The problem of determining path numbers appears to have its motivation in the Canvas protocol for wireless sensor networks [15]. Other applications emerge in the context of security of computer networks [1] and processing of large graphs repre- senting road networks (eg. road planning), see [9]. In all these areas in addition to determining k-PVCs and ψk(G) for a fixed k, it is also natural to ask another question: given a number t of vertices (representing tamper-proof sensors, firewalls in a network, POIs on a map, etc.) find path lengths for which t is sufficient in terms of system security, map accuracy, etc. In other words, find the largest number k such that ψk(G) > t. One can also consider ‘upgrades’ of the system in question, for ex- ample: if we want to decrease k (i.e. increase system security, increase map accuracy, etc.), how much needs t be increased? These questions are clearly related to a path sequence and its subsequences.
Path numbers generalize some well-known problems from graph theory, for exam- ple the cardinality of a minimum vertex cover of a graph G equals ψ2(G), dissociation number of G is equal to |V | − ψ3(G) (see [3], [12]) and in general values of ψk(G) are exactly cardinalities of minimum vertex covers of k-uniform ’path hypergraph’
H built from G (every path P on k vertices in G gives rise to a hyperedge in H containing vertices of P , see [3]). The task of computing ψk(G) is NP-complete for a general graph G and any k ≥ 2 (one can use a reduction from minimum vertex cover problem given in [3], Theorem 1). However, there do exist classes of graphs for which polynomial-time algorithms are available for all k - for instance trees as shown by Breˇsar et al. in [3] or graphs with bounded tree-width (see [3], [6]). Apart from that, exact formulae to compute ψk(G) are known only for some special classes of graphs and particular values of k. These include for example grids and k = 3 (see [4]), rooted product graphs and k = 3 (see [10]). In the case of other graph products some bounds and exact values (for k = 2, 3) are also given in [11]. The paper [10]
gives one of the very few known relations between the numbers ψm(G) for different m (namely between the value of ψk(G◦ H) in a product graph and values of ψk(G), ψk(H) and ψ2(H) of its factors) – in Section 3. we present some new results of this kind.
As for the approximation of path numbers, following [3] one can adapt standard 2-approximation algorithm for the minimum vertex cover problem to approximate ψk(G) within a factor of k – but polynomial-time approximation with better constant factor is not known in general for ψk(G) and k > 3. For k = 3 however, there exist algorithms of Tu and Zhou [17], [18] with an approximation factor of 2. Some recent results for approximating ψ3(G) and ψ4(G) in cubic graphs are also given in [13], [16].
On the other hand, the famous result of Dinur and Safra [8] shows that it is NP-hard to approximate the value of ψ2(G) to within any factor smaller than 10√
5−21 = 1.3606...
(unless P = NP). In [7] it is shown that approximation of minimum vertex cover in k-uniform hypergraphs for k≥ 3 is NP-hard within a constant factor below k − 1 and thus it seems unlikely that better approximation of ψk(G) can be achieved. Bearing that in mind, it appears difficult to obtain a constant approximation factor for a path sequence as a whole (i.e. approximate all ψk(G) within the same factor, independent of k and number of vertices in G), but the problem is worth further research.
2. Elementary results
Unless otherwise stated, in the following G will denote a (simple, nonempty) graph on n vertices and k a positive integer satisfying 1≤ k ≤ n.
For the standard notations and definitions in graph theory we refer the reader to [2]; here we only recall some extensively used notations. If G = (V, E) is a graph and S⊂ V , then G[S] denotes the subgraph induced by S. Now for v ∈ V and e ∈ E we denote by G− v the graph G[V \ {v}] and by G − e the graph (V, E \ {e}). We will also write|G| for the number of vertices in G. By Pn, Cn and Kn we denote a path, a cycle and a complete graph on n vertices respectively. A complete bipartite graph with partitions of size a and b will be denoted by Ka,b. The symbol' denotes graph isomorphism and by ’disjoint graphs’ we mean vertex disjoint graphs. For a vertex v of G we denote by d(v) the degree of v and by N (v) the neighbourhood of v (the set of all vertices adjacent to v).
Let us first note that from the definition of path numbers one immediately gets ψ1(G) = n, ψk(G)≤ n − k + 1 (an arbitrary subset of n − k + 1 vertices is a k-PVC) and that a path sequence is non-increasing, i.e. ψ1(G) ≥ ψ2(G) ≥ ψ3(G) ≥ . . . ≥ ψn(G)≥ 0. An easy calculation gives path numbers for paths, cycles and complete graphs, namely ψk(Pn) =n
k
, ψk(Cn) =n
k
and ψk(Kn) = n− k + 1 (see [4]). We present values for complete bipartite graphs below.
Proposition 2 Let 1≤ k ≤ a + b. Then:
ψk(Ka,b) =
a + b if k = 1,
min{a, b} −k 2
+ 1 for 1 < k≤ 2 min{a, b} + 1,
0 otherwise.
Proof. Let us write A and B for partitions of Ka,b with|A| = a ≤ b = |B|. The case k = 1 is clear. Assume that 1 < k≤ a + b and take p = a −k
2
+ 1. Since any path in Ka,b alternates between A and B, a path on k vertices must have at leastk
2
vertices in common with A. It follows that for k > 2a + 1 there is no path on k vertices in Ka,b and so ψk(Ka,b) = 0. So let k≤ 2a + 1 and note that from the above reasoning an arbitrary set of p vertices in A is a k-PVC and consequently ψk(Ka,b)≤ p.
Now let T be a subset of A∪ B and |T | = p − 1. To show that ψk(Ka,b) > p− 1 we will build a path on k vertices disjoint from T . It suffices to find an arbitrary set ofk
2
vertices in A\ T andk
2
vertices in B\ T (or vice versa). Note that there are at least a− p + 1 =k
2
vertices in A\ T and also at least b − p + 1 ≥k
2
vertices in B\ T , since b ≥ a. If b > a, then b − p + 1 ≥k
2
+ 1≥k
2
and the result follows. So we can assume that a = b. If|A \ T | >k
2
, then|A \ T | ≥k
2
and we are done since
|B \ T | ≥k
2
. By symmetry, the only case left is|A| = |B|, |A \ T | = |B \ T | =k
2
. But if|A \ T | =k
2
, then T ⊂ A and B ∩ T = ∅, so |B \ T | = |B|. It follows that
|A \ T | = |B \ T | = |B| = |A| and so T = ∅; consequently p = 1 and a =k
2
, so k = 2a or k = 2a + 1. It is easily verified that ψ2a(Ka,a) = 1 and the value agrees with the formula given in the proposition; the case k = 2a + 1 is impossible since k≤ a + b = 2a.
Let us also note the following useful lemma:
Lemma 3 Let G = (V, E) be a graph on n≥ 2 vertices, k < n and v ∈ V . Then ψk(G)≤ ψk(G− v) + 1. Moreover, the following conditions are equivalent:
1. ψk(G) = ψk(G− v) + 1,
2. ∃S ⊂ V : S is a minimum k-PVC for G and v ∈ S,
3. ∃T ⊂ V \ {v} : T is a k-PVC for G − v and T ∪ {v} is a minimum k-PVC for G.
Proof. Let U be a minimum k-PVC for G− v. Then clearly U ∪ {v} is a k-PVC for G and so ψk(G)≤ ψk(G− v) + 1.
Now suppose that (1) holds and U is as above - then U∪ {v} is a minimum k-PVC for G and (2) follows with S = U∪ {v}. If (2) holds, then T = S \ {v} is a k-PVC for G− v (a path disjoint from T in G − v is disjoint from S in G) and (3) clearly follows. Supposing that (3) holds gives ψk(G) =|T | + 1 ≥ ψk(G− v) + 1 ≥ ψk(G) by the first part of the lemma, so ψk(G) = ψk(G− v) + 1.
As a corollary we get:
Corollary 4 Let G = (V, E) be a graph and e∈ E. Then ψk(G)≤ ψk(G− e) + 1.
Proof. Let e = uv. Since G− u is a subgraph of G − e, we apply previous lemma to obtain ψk(G)≤ ψk(G− u) + 1 ≤ ψk(G− e) + 1.
The following remark shows that there are no restrictions on the structure of a minimum k-PVC.
Remark 5 For any graph H = (W, F ), there exists a supergraph G of H such that G[W ]' H and that W is a minimum k-PVC for G.
Proof. We adapt the construction from the proof of [3], Theorem 1. So let us replace each vertex v of H with a path P(v)= v− w(v)2 − . . . − w(v)k on k vertices (paths for different v are pairwise disjoint), leaving edges F intact. Call the resulting graph G and note that G[W ] ' H. Now any k-PVC for G must contain at least one vertex from each path P(v), so ψk(G)≥ |W |. But W is clearly a k-PVC, so ψk(G) =|W |.
3. Two element subsequences
In this section we investigate relations between two path numbers ψk(G) and ψm(G) for an arbitrary graph G. Let us start with the following example, showing that two elements of a path sequence must satisfy some additional conditions apart from the ones given in the previous section.
Example 6 There is no graph G satisfying ψ10(G) = 2 and ψ2(G) = 5.
Proof. Suppose that such a graph G exists. Let S be a minimum 2-PVC for G and v ∈ S. Note that at least one vertex from any edge in G belongs to S. Take an arbitrary path P in G that avoids v. Since for every two consecutive vertices on P at least one is from S and P has no more than 4 vertices in common with S – it follows that P is a path on at most 9 vertices. This shows that{v} is a 10-PVC for G and so ψ10(G)≤ 1, a contradiction.
These additional necessary conditions are presented in the following proposition.
Proposition 7 Let 1≤ m < k and ψk(G) > 0. Then ψm(G)≥ ψk(G) +k
m
− 1.
Proof. We will proceed by induction on n - the number of vertices in G. The result clearly follows for n≤ 2 and also for all graphs G with ψk(G) = 1 (including the case k = n), because we have ψm(G)≥ ψm(Pk) =k
m
. So we can assume that n > k≥ 2 and ψk(G) > 1. Let S be a minimum m-PVC for G and v∈ S. By Lemma 3, we get ψm(G) = ψm(G−v)+1 and ψk(G−v) ≥ ψk(G)−1 > 0. By the induction hypothesis ψm(G− v) ≥ ψk(G− v) +k
m
− 1 and consequently ψm(G) = ψm(G− v) + 1 ≥ ψk(G− v) +k
m
≥ ψk(G)− 1 +k
m
.
Remark 8 Notice that the condition ψk(G) > 0 in the above proposition cannot be omitted: take for example G = K1,8 (a star on 9 vertices), k = 9 and m∈ {2, 3, 4}.
Remark 9 Let m < k and take G equal to s disjoint copies of Pm. Then clearly ψm(G) = s and ψk(G) = 0. This shows that there exist graphs G with ψk(G) = 0 and an arbitrary value of ψm(G).
As the converse of Proposition 7 we show the following:
Theorem 10 Let k be a positive integer and 1 ≤ m < k. If two integers pk, pm
satisfy: pk> 0 and pm≥ pk+k
m
− 1, then there exists a (connected) graph G such that ψk(G) = pk and ψm(G) = pm.
Proof. Let k
m
= a, i.e. am ≤ k < (a + 1)m. Take pk + a− 1 disjoint paths P(1), P(2), . . . , P(pk+a−1)on 2m−1 vertices and let P(i)= v(i)1 −v2(i)−. . .−v2m(i)−1. Now add edges connecting vertices v(i)x and v(j)m for all i6= j and all x, i.e. 1 ≤ x ≤ 2m − 1.
Call the resulting graph H and let M ={vm(i): 1≤ i ≤ pk+ a− 1} denotes the set of “middle” vertices of all P(i) (see Fig. 1). Since ψm(P(i)) = 1, it is easily seen that M is a minimum m-PVC for H and so ψm(H) =|M| = pk+ a− 1.
Let us now prove the following lemma concerning H and M :
Lemma 11 Any path on k vertices in H must contain at least a vertices from M . Proof. Suppose to the contrary that W = w1−. . .−wkis a path in H and|{w1, . . . , wk}
∩M| = t < a. Let us divide W into consecutive fragments (subpaths) contained in paths P(i), i.e. W = W1− . . . − Ws, where each Wj is a (maximal) subpath of W with all vertices in some fixed P(i). Note that it is possible for multiple Wj to be subpaths of a single P(i)and that any Wj with at least m vertices must contain some vertex from M . Let us now define two sets of indices j: I0 = {j : Wj ∩ M = ∅}
and I1 ={j : Wj∩ M 6= ∅} = {j : Wj has exactly one vertex in common with M}.
Clearly, the total number of Wj equals s =|I0| + |I1| = |I0| + t. Now we write I1
Figure 1. An example graph H for m = 3 and pk+ a− 1 = 3. Paths P(i)are drawn horizontal, set M is marked in black.
as a disjoint sum of the following subsets: B ={j : Wj has more than m vertices}, U ={j : Wj is a single vertex from M} and R = I1\ (B ∪ U) = {j ∈ I1: 2≤ |Wj| ≤ m}. Counting the number of vertices in W as a sum of the numbers of vertices in Wj yields the following bound:
|W | = X
j∈I0∪I1
|Wj| = X
j∈I0∪B∪U∪R
|Wj| ≤
≤|I0| · (m − 1) + |B| · (2m − 1) + |U| · 1 + |R| · m =
=|I0| · (m − 1) + |B| · (2m − 1) + |U| + (s − |I0| − |B| − |U|) · m =
=(s +|B| − |U|) · m + |U| − |I0| − |B|
We will now show the following claim: let a < b be two integers such that the last (in order imposed by W ) vertex in Waand the first vertex in Wb are not in M . Then there exists an index p∈ U such that a < p < b. Indeed, by the construction of H and Wj, since the last vertex in Wa is not in M , the first one in Wa+1 must be in M . Analogously, the last vertex in Wb−1 must be in M . If a + 1∈ U or b − 1 ∈ U, then we are done. If not, the last vertex in Wa+1 and the first one in Wb−1 are not in M and we can proceed by induction on b− a (the case b − a = 1 being impossible and b− a = 2 easily verified).
Since any a, b ∈ I0∪ B such that a < b satisfy the hypothesis of the claim, we get |U| ≥ |I0| + |B| − 1. The bound for |W | attains its maximum for the smallest possible|U|, that is |U| = |I0| + |B| − 1 and then we get |W | ≤ (s − |I0| + 1) · m − 1 = (t + 1)· m − 1 ≤ am − 1 < k, a contradiction that ends the proof.
We will show that ψk(H) = pk. First note that by Lemma 11 we get ψk(H)≤ pk, since the set S = {vm(i) : 1 ≤ i ≤ pk} is a k-PVC for H as there are only a − 1 vertices in M\ S. Now let T be an arbitrary set of no more than pk− 1 vertices from H. Without loss of generality we can assume that T has no vertices in common
with paths P(1), . . . , P(a) (recall that the number of P(i)is pk+ a− 1). It is easy to observe that joining the paths v(1)1 − . . . − v2m(1)−1, v(2)m − vm+1(2) − . . . − v2m(2)−1, . . . , vm(a)− v(a)m+1− . . . − v(a)2m−1 results in a path on 2m− 1 + (a − 1)m = (a + 1)m − 1 ≥ k vertices. Consequently T is not a k-PVC for H and ψk(H) > pk− 1, so we must have ψk(H) = pk.
Now we deal with the m-path number. Take pm− (pk+ a− 1) (by assumption this number is non-negative) disjoint paths Q(1), Q(2), . . . , Q(pm−(pk+a−1)) on m vertices and let Q(j)= u(j)1 − . . . − u(j)m. Connect all Q(j)to the vertex v(1)m of P(1)by adding edges u(j)1 − v(1)m for all j. Resulting graph G satisfies ψk(G) = ψk(H) = pk, since S ={vm(i): 1≤ i ≤ pk} is a k-PVC for G. But also ψm(G) = ψm(H) + (pm− (pk+ a− 1)) = pmbecause at least one vertex from each P(i)and each Q(j) must be included in the minimum m-PVC of G and clearly M∪ {u(j)1 : j = 1, 2, . . . , pm− (pk+ a− 1)}
is a m-PVC for G.
4. Path sequences for small graphs
In this section we give some properties of path sequences concerning graphs with small number of vertices. First problem which arises naturally is the question whether equality of path sequences implies graph isomorphism. This is true for graphs with at most three vertices but false in general, as shown by the proposition below.
Proposition 12 For any n≥ 4 there exist (connected) graphs G, H on n vertices such that ψ(G) = ψ(H) but G and H are not isomorphic.
Proof. Let G be a graph and v a vertex in G. By Gv,k we understand a graph obtained from G by adding k new vertices{u1, . . . , uk} and edges {vu1, vu2, . . . , vuk} to G. Now, let u∈ V (C4) and let v∈ V (K4− e) be of degree 3 and consider graphs G = (C4)u,n−4 and H = (K4− e)v,n−4. Obviously G and H are not isomorphic but
ψ(G) = ψ(H) =
(4, 2, 2, 1) if n = 4, (n, 2, 2, 1, 1, 0, . . . , 0) for n≥ 5.
Before going further we state the following definition:
Definition 13 Let (p1, . . . , pn) be a sequence of non-negative integers. Put m(p1, . . . , pn) := number of non-isomorphic connected graphs G
on n vertices such that ψ(G) = (p1, . . . , pn).
We will call this number the path multiplicity of a sequence (p1, . . . , pn). A sequence with nonzero path multiplicity will be called realisable, i.e. (p1, . . . , pn) is realisable if there exists a connected graph G with ψ(G) = (p1, . . . , pn). Moreover if at least one of the graphs G satisfying ψ(G) = (p1, . . . , pn) is a tree, a bipartite graph, etc. we will say that the sequence is realisable by a tree, a bipartite graph, etc.
Tables 1 and 2 give realisable sequences and their path multiplicities for connected graphs on n = 5, 6 and 7 vertices (for smaller n all values are easily found ‘by hand’).
These numbers were generated using a computer program written by the authors – source code is available at [20]. The lists of non-isomorphic connected graphs and trees were obtained by the Mathematica package [19] and data from the web page [14]. Note: sequences realisable by trees are marked with∗.
Table 1. Path sequences for all connected graphs on 5 vertices (9 sequences, 21 graphs) and 6 vertices (20 sequences, 112 graphs).
Sequence Multiplicity
∗(5,1,1,0,0) 1
∗(5,2,1,1,0) 2
∗(5,2,1,1,1) 1 (5,2,2,1,1) 5 (5,3,1,1,1) 2 (5,3,2,1,1) 2 (5,3,2,2,1) 5 (5,3,3,2,1) 2 (5,4,3,2,1) 1
Sequence Mult. Sequence Mult.
∗(6,1,1,0,0,0) 1 (6,3,3,2,1,1) 5
∗(6,2,1,1,0,0) 2 (6,3,3,2,2,1) 14
∗(6,2,2,1,0,0) 1 (6,4,2,1,1,1) 4
∗(6,2,2,1,1,0) 10 (6,4,2,2,1,1) 1
∗(6,3,1,1,1,0) 3 (6,4,2,2,2,1) 8 (6,3,2,1,1,0) 3 (6,4,3,2,1,1) 2
∗(6,3,2,1,1,1) 9 (6,4,3,2,2,1) 7 (6,3,2,2,1,0) 1 (6,4,3,3,2,1) 9 (6,3,2,2,1,1) 22 (6,4,4,3,2,1) 3 (6,3,2,2,2,1) 6 (6,5,4,3,2,1) 1
Table 2. Path sequences for all connected graphs on 7 vertices (50 sequences, 853 graphs).
Sequence Multiplicity Sequence Mult. Sequence Mult.
∗(7,1,1,0,0,0,0) 1 (7,3,3,2,2,1,0) 1 (7,4,4,3,2,1,1) 6
∗(7,2,1,1,0,0,0) 2 (7,3,3,2,2,1,1) 87 (7,4,4,3,2,2,1) 24
∗(7,2,2,1,0,0,0) 1 (7,4,1,1,1,0,0) 3 (7,4,4,3,3,2,1) 36
∗(7,2,2,1,1,0,0) 16 (7,4,2,1,1,1,0) 6 (7,5,3,1,1,1,1) 3
∗(7,3,1,1,1,0,0) 4 (7,4,2,1,1,1,1) 3 (7,5,3,2,1,1,1) 4
∗(7,3,2,1,1,0,0) 5 (7,4,2,2,1,1,0) 1 (7,5,3,2,2,1,1) 1
∗(7,3,2,1,1,1,0) 21 (7,4,2,2,1,1,1) 25 (7,5,3,2,2,2,1) 18
∗(7,3,2,1,1,1,1) 2 (7,4,2,2,2,1,1) 39 (7,5,3,3,2,1,1) 1 (7,3,2,2,1,0,0) 1 (7,4,3,1,1,1,0) 1 (7,5,3,3,2,2,1) 8 (7,3,2,2,1,1,0) 39 (7,4,3,1,1,1,1) 12 (7,5,3,3,3,2,1) 22 (7,3,2,2,1,1,1) 4 (7,4,3,2,1,1,0) 3 (7,5,4,3,2,1,1) 2 (7,3,2,2,2,1,0) 1 (7,4,3,2,1,1,1) 20 (7,5,4,3,2,2,1) 7 (7,3,2,2,2,1,1) 9 (7,4,3,2,2,1,1) 69 (7,5,4,3,3,2,1) 19 (7,3,3,1,1,1,0) 3 (7,4,3,2,2,2,1) 81 (7,5,4,4,3,2,1) 15 (7,3,3,1,1,1,1) 8 (7,4,3,3,2,1,1) 46 (7,5,5,4,3,2,1) 3 (7,3,3,2,1,1,0) 10 (7,4,3,3,2,2,1) 129 (7,6,5,4,3,2,1) 1
(7,3,3,2,1,1,1) 10 (7,4,3,3,3,2,1) 20 - -
From Tables 1 and 2 we can draw some observations. First of all notice that from the basic properties of path numbers it follows that m(n, n− 1, n − 2, . . . , 1) = 1 (realisable by Kn) and that m(n, 1, 1, 0, . . . , 0) = 1 (realisable by K1,(n−1), for n≥ 3).
However there are many other path sequences with multiplicity one, for example m(5, 2, 1, 1, 1) = 1.
Proposition 12 shows that equality of path sequences for two graphs does not imply that they are isomorphic. The tables above show that we can even have that ψ(G) = ψ(T ) for some tree T and some non-tree graph G.
However if T1, T2 are trees with n < 7 vertices, then it follows from Tables 1 and 2 (and the number of non-isomorphic trees on n vertices) that ψ(T1) = ψ(T2) ⇐⇒
T1' T2. But this is also not true in general, as the following proposition shows.
Proposition 14 For any n ≥ 7 there exist trees T1, T2 on n vertices such that ψ(T1) = ψ(T2) but T1 and T2 are not isomorphic.
Proof. Assume first that n = 7 and take the following trees:
W
T1
W
T2
that are clearly not isomorphic and have path sequences equal to (7, 2, 2, 1, 1, 0, 0). For n > 7 it suffices to attach additional vertices to w (consider (T1)w,n−7 and (T2)w,n−7
in notations of Proposition 12) to obtain non-isomorphic trees with path sequences (n, 2, 2, 1, 1, 0, . . . , 0).
By analysing the data for graphs with up to 9 vertices (path sequences were cal- culated with the help of the computer program [20]) we state the following conjecture concerning the existence of a Hamilton path in G (that is clearly equivalent to the condition ψn(G) = 1). According to our knowledge this conjecture has not been studied yet.
Conjecture 15 Let G be a connected graph on n≥ 2 vertices. Then the following implication holds
ψn−1(G) = 2⇒ ψn(G) = 1
By setting k = n− 1 in the following remark, one observes that it is not necessary to formulate Conjecture 15 for disconnected graphs.
Remark 16 If G is a graph on n vertices such that ψk(G) = n− k + 1 for some 2≤ k ≤ n, then G is connected.
Proof. Let n≥ 2 and let Hi, i∈ I be the connected components of G. Since ψk(Hi)≤
|Hi| − k + 1, we get ψk(G) =P
i∈Iψk(Hi)≤P
i∈I(|Hi| − k + 1) = n − (k − 1) · |I|.
Now if ψk(G) = n− k + 1, then n − k + 1 ≤ n − (k − 1) · |I|. Since k ≥ 2, equality is possible only for|I| = 1, i.e. when G is connected.
It is straightforward to see that if G is a graph on n vertices and ψ2(G) is maximum possible (i.e. n− 1), then G is necessarily isomorphic to Kn. It is not the case for
ψk(G) and k > 2, however Conjecture 15 implies the following interesting property of path sequences.
Theorem 17 Let G be a graph on n≥ 3 vertices and 2 ≤ k < n.
If Conjecture 15 holds for all connected graphs with at most n vertices, then ψk(G) = n− k + 1 implies ψj(G) = n− j + 1 for all j such that k < j ≤ n.
Proof. It is enough to prove the following claim for all graphs G on n≥ 3 vertices and all m such that 2≤ m < n: if Conjecture 15 holds for all connected graphs with at most n vertices, then ψm(G) = n− m + 1 implies ψm+1(G) = n− m.
We proceed by induction on n. It easy to check the claim for n = 3. So assume that n≥ 4 and fix m ∈ {2, 3, . . . , n − 1}. Let G be a graph on n vertices such that ψm(G) = n− m + 1 and ψm+1(G) = n− m − t, with some t ≥ 0. We need to prove that t = 0. Observe that G is connected by Remark 16 and if m = n− 1, then the result follows by the validity of Conjecture 15. So assume that m < n− 1 and put S to be a minimum (m + 1)-PVC for G. There are two cases to consider:
(a) S6= ∅. Choose v ∈ S and let G0 = G− v. Lemma 3 gives n − m ≤ ψm(G0)≤ (n − 1)−m+1 and therefore ψm(G0) = n−m. By the induction hypothesis ψm+1(G0) = n− m − 1, but due to Lemma 3 we obtain that ψm+1(G0) = ψm+1(G)− 1 = n− m − t − 1, so t = 0.
(b) S =∅. This case cannot occur and the proof is as follows: let w be any vertex of G, put G0= G−w and observe that by Lemma 3 we get ψm(G0) = n−m, so by the induction hypothesis ψm+1(G0) = n− m − 1. But 0 ≤ ψm+1(G0)≤ ψm+1(G) = 0 and consequently n− m − 1 = 0, which is impossible since m < n − 1.
We now present a lemma which will be useful in giving a direct proof of Conjecture 15 for graphs with no more than 7 vertices. Note that the points (2)–(4) follow from Lemma 2.2 of [5], however we include them here with a proof.
Lemma 18 Let G = (V, E) be a connected graph on at least four vertices such that V ={p1, . . . , pn−1, q}, N(q) = {pi1, . . . , pit} and p1− . . . − pn−1 is a path in G. If ψn−1(G) = 2 and ψn(G) = 0, then
(1) d(p1)≥ 2 and d(pn−1)≥ 2.
(2) p1pn−1 ∈ E, qp/ 1 ∈ E, qp/ n−1 ∈ E and if qp/ i ∈ E, then qpi+1 ∈ E for all/ i∈ {2, . . . , n − 2}.
(3) p1pij+1∈ E and p/ n−1pij−1∈ E for all j ∈ {1, . . . , t}./
(4) p1pij−1∈ E, for all j such that i/ j> min{i1, . . . , it} and pn−1pij+1∈ E for all j/ such that ij< max{i1, . . . , it}.
Proof. To see (1) suppose to the contrary that d(p1) = 1. We show that S ={p2} is a (n− 1)-PVC for G. This follows from the fact that any path on k vertices which avoids p2 must also avoid p1 and therefore k ≤ n − 2. The second case is proved analogously.
Now, the first part of (2) is obvious (in any case we get a path on all vertices in G, which contradicts ψn(G) = 0). For the second, if such an i exists, we have a path p1− . . . − pi− q − pi+1− . . . − pn−1.
As far as (3) is concerned, suppose first that p1pij+1 ∈ E for some ij such that pij ∈ N(q). Then we have the following path on n vertices in G: q − pij − pij−1− . . .− p1− pij+1− pij+2− . . . − pn−1. If now pn−1pij−1 ∈ E, then we get that the following path on n vertices: p1− . . . − pij−1− pn−1− . . . − pij − q exists.
To prove (4), let r = min{i1, . . . , it} and suppose that p1pij−1∈ E. Then we have the following path on n vertices in G: pn−1− pn−2− . . . − pij − q − pr− pr−1− . . . − p1− pij−1− pij−2− . . . − pr+1. The second case follows by symmetry argument.
Corollary 19 Let G = (V, E) be a connected graph on at least four vertices such that V ={p1, . . . , pn−1, q} and p1− . . . − pn−1 is a path in G. If ψn−1(G) = 2 and ψn(G) = 0, then
2≤ d(q) ≤
n− 3 2
Proof. Firstly notice that d(q)≥ 2, since if d(q) = 1 and qu ∈ E, then S = {u} is a (n− 1)-PVC. To see this note that any path P in G that avoids u must also avoid q and so|P | < n − 1. As for the second inequality, it follows from Lemma 18.(2) since for every i∈ {2, . . . , n − 2} at most one of the edges qpi, qpi+1 is in E.
The above facts allow us to give a direct proof of Conjecture 15 for graphs with no more than 7 vertices. Our reasoning is ‘by considering cases’ – unfortunately we were unable to find a more general approach.
Theorem 20 Let G = (V, E) be a connected graph on n vertices, with 2≤ n ≤ 7.
Then Conjecture 15 holds for G, i.e.
ψn−1(G) = 2⇒ ψn(G) = 1
Proof. The theorem holds true for n = 2 and n = 3, with complete graphs K2 and K3 being the only cases to verify. So we can assume that n≥ 4.
It is sufficient to prove that the existence of a connected graph that satisfies ψn−1(G) = 2 and ψn(G) = 0 leads to a contradiction. Throughout we consider a graph G = (V, E) with V ={p1, . . . , pn−1, q} and assume that p1− . . . − pn−1 is a path in G.
Notice that if n = 4 or n = 5, thenn−3
2
< 2 and Corollary 19 gives 2≤ d(q) < 2, a contradiction. So we can assume that n = 6 or n = 7. Note that in both cases we get d(q) = 2 by Corollary 19.
Now let G be a connected graph with 6 vertices such that ψ5(G) = 2, ψ6(G) = 0 and d(q) = 2. Due to Lemma 18 we only need to consider the case when qp2, qp4∈ E - but then the set S5={p2} is a 5-PVC. To see this let us assume that there exists a path P on 5 vertices p1, p3, p4, p5, q (in any order). Using again Lemma 18 we obtain that d(p1) = 2,p1p4∈ E and p3p5∈ E. It follows that p/ 2p3 and p3p4are the only two edges containing p3. Consequently P must start with p3and it is easy to see that we cannot build a path avoiding p2 longer than p3− p4− x where x ∈ {q, p1, p5}, which contradicts P having 5 vertices.
Let us now assume that n = 7 and there exists a graph G such that ψ6(G) = 2 and ψ7(G) = 0 with d(q) = 2. Because of Lemma 18 and symmetries we only need to consider two cases: qp2, qp5 ∈ E and qp2, qp4 ∈ E. Assume the first case – by Lemma 18 we get that d(p1) = 2 and p1p5∈ E. But now S6={p2} is a 6-PVC for
G: this follows from the fact that there is no path on 6 vertices which avoids p2in G.
Indeed, if such a path exists, then it is of the form q− p5− x1− x2− x3− x4 where xi6= p1for i = 1, 2, 3, 4 – a contradiction.
Let us now proceed with the case qp2, qp4∈ E. By Lemma 18 we must have that d(p1) = 2 and p1p4 ∈ E. But then again S6 ={p2} is a 6-PVC for G. To see this, notice that we only need to consider paths of the form q− p4− x1− x2− x3− x4. If such a path omits p2, then we cannot have x1= p1and so we get xi ∈ {p3, p5, p6} for i = 1, 2, 3, 4 – a contradiction.
As a consequence of the above and Theorem 17 we get
Corollary 21 Let G be a graph on n vertices with 3≤ n ≤ 7 and let 2 ≤ k < n. If ψk(G) = n− k + 1, then ψj(G) = n− j + 1 for all j such that k < j ≤ n.
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